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One of the problems in the text book was asking us to determine if

$$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n^2}{(n+1)^2}$$

would converge absolutely, conditionally, or diverge. The way I found my solution was different from what we did in class. I started listing out terms:

$$\frac14,-\frac49,\frac{9}{16},-\frac{16}{25},\dots$$

and I grouped every two terms as one term (for example $1/4-4/9$ would be the $n=1$ term).

Using this grouping technique I formed what seemed like an equivalent series:

$$\sum_{n=1}^{\infty}\frac{(2n-1)^2}{4n^2}-\frac{4n^2}{(2n+1)^2}.$$

Before I had changed the series to this form all the tests for convergence had failed, but in this new form, after combining it into one fraction and using limit comparison test with $1/n^2$, it appears to converge. The back of the book says it diverges but my method says otherwise. Since official documents are always right I want to know what was wrong in my ideology.

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    Did you try the term test? It's the first test you should try. – 2017-02-23
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    Even though it is a alternating series, I used the term test and the nth term was 1 which would indicate divergence, but my real question is what is wrong about my other method that makes it predict convergence? – 2017-02-23
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    It is possible to rearrange some diverging series to turn them into a convergent series. It is like saying this: $$1-1+1-1+\dots=(1-1)+(1-1)+\dots=\text{converges}$$ – 2017-02-23

2 Answers 2

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Notice that for your initial series, the general term does not approach $0$ and hence it diverges.

Regrouping terms as you did is generally not allowed. For instance, the series

$$\sum_{n\geq 1}(-1)^n$$

does not converge in any classical sense -- its partial sums oscilate between $-1$ and $0$ --, but with your 'method' it does. What you did is produce a new series, and convergence of that new series does not imply convergence of the old one.

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Always remember that the convergence of a series is defined as the convergence of the sequence of partial sums. It is not enough to have a sequence of some of the partial sums converging; you must have the entire sequence of partial sums converging. In your case, by combining terms in pairs, you are effectively taking only every $(2k)$-th partial sum for natural number $k$. This is exactly the same phenomenon that you can see in the divergent sequence $((-1)^n)_{n\in\mathbb{N}} = (1,-1,1,-1,\cdots)$, where if you take the even-indexed terms you get a convergent sequence $(1)_{n\in\mathbb{N}} = (1,1,1,1,\cdots)$.